How do we evaluate ∫1+x2(1−x2)1+x4−−−−−√dx? ∫ 1 + � 2 ( 1 − � 2 ) 1 + � 4 d � ?

 𝙰𝙽𝚂𝚆𝙴𝚁 = One possible way to evaluate the integral ∫(1+x^2(1−x^2)1+x^4)^(-1/2)dx is by using trigonometric substitution. Let x = tanθ, so that dx = sec^2θdθ and

1 + x^2 = 1 + tan^2θ = sec^2θ

1 - x^2 = 1 - tan^2θ = cos^2θ

1 + x^4 = 1 + tan^4θ = sec^4θ

Substituting these expressions in the integral yields:

∫(1+x^2(1−x^2)1+x^4)^(-1/2)dx = ∫(sec^2θcos^2θsec^4θ)^(-1/2)sec^2θdθ

= ∫(cos^2θ)^(1/2)(sec^4θ)^(1/2)sec^2θdθ

= ∫cosθsec^3θdθ

We can evaluate this integral using integration by substitution:

Let u = secθ, so that du = secθtanθdθ

Substituting u and du, we have:

∫cosθsec^3θdθ = ∫du/u^3 = -1/(2u^2) + C

Substituting back u = secθ, we get:

∫cosθsec^3θdθ = -1/(2sec^2θ) + C

= -1/(2+2x^2(1-x^2)^(-1/2)) + C

= -1/(2+2tan^2θsecθ) + C

= -1/(2+x√(1-x^2)) + C

Therefore, the solution to the integral is:

∫(1+x^2(1−x^2)1+x^4)^(-1/2)dx = -1/(2+x√(1-x^2)) + C.


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